Explanation

(b)
Let the straight line meets the $x$-axis at $A(a,0)$ and the $y$-axis at $B(0,\ b)$. The equation of this straight line will be
$\frac{x}{a} + \frac{y}{b} = 1$ …(i)
Since, it passes through $P(1,\ 1)$
$\therefore\frac{1}{a} + \frac{1}{b} = 1\ \Rightarrow \ a + b = ab$ …(ii)
Let the coordinates of the mid point $M$ of $AB$ are $(h,k)$
$\therefore\ h = \frac{a + 0}{2}\ \Rightarrow \ a = 2h$
and $k = \frac{0 + b}{2} \Rightarrow b = 2k$
substitute the values of $a$ and $b$ in Eq. (ii), we get
$2h + 2k = 2h \times 2k$
$\Rightarrow \ h + k = 2hk$
Hence, the equation of the locus of mid point $M(h,k)$ will be
$x + y - 2xy = 0$